Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long int maxn = 2e5 + 10, inf = 3e18, mod = 1e9 + 7, LOG = 30;
long long int n, m, q, h[maxn], par[maxn][LOG];
vector<int> a[maxn], vec[maxn][LOG], adj[maxn];
void dfs(int u, int v) {
  for (int i : adj[u])
    if (i != v) par[i][0] = u, h[i] = h[u] + 1, dfs(i, u);
}
inline vector<int> merge(vector<int> v1, vector<int> v2) {
  int i, j, cnt;
  vector<int> v;
  i = j = cnt = 0;
  while (cnt < 10 && (i < v1.size() || j < v2.size())) {
    if (i < v1.size() && j < v2.size()) {
      if (v1[i] < v2[j]) {
        v.push_back(v1[i]);
        i++;
      } else {
        v.push_back(v2[j]);
        j++;
      }
      cnt++;
    } else if (i < v1.size()) {
      cnt++;
      v.push_back(v1[i]);
      i++;
    } else {
      v.push_back(v2[j]);
      j++;
      cnt++;
    }
  }
  return v;
}
vector<int> lca(int u, int v, int num) {
  vector<int> mns;
  if (u == v) {
    mns = merge(mns, a[u]);
    while (mns.size() > num) mns.pop_back();
    return mns;
  }
  if (h[u] < h[v]) swap(u, v);
  int log = LOG - 1;
  if (u != v) {
    mns = merge(mns, a[u]);
  }
  while (h[u] != h[v] && log >= 0) {
    if ((1 << log) <= h[u] - h[v]) {
      mns = merge(mns, vec[u][log]);
      u = par[u][log];
    }
    log--;
  }
  if (u == v) {
    while (mns.size() > num) mns.pop_back();
    return mns;
  }
  mns = merge(mns, a[v]);
  sort(mns.begin(), mns.end());
  log = LOG - 1;
  while (log >= 0) {
    if (par[u][log] != par[v][log]) {
      mns = merge(mns, vec[u][log]);
      mns = merge(mns, vec[v][log]);
      u = par[u][log], v = par[v][log];
    }
    log--;
  }
  u = par[u][0];
  mns = merge(mns, a[u]);
  while (mns.size() > num) mns.pop_back();
  return mns;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int test;
    cin >> test;
    a[test].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < LOG; j++) par[i][j] = i;
  dfs(1, 1);
  for (int j = 1; j < LOG; j++)
    for (int i = 1; i <= n; i++) par[i][j] = par[par[i][j - 1]][j - 1];
  for (int j = 0; j < LOG; j++) {
    for (int i = 1; i <= n; i++) {
      if (h[i] >= (1 << j)) {
        if (j == 0) {
          if (i == 1) continue;
          vec[i][j] = a[par[i][0]];
          continue;
        }
        vec[i][j] = merge(vec[i][j - 1], vec[par[i][j - 1]][j - 1]);
      }
    }
  }
  while (q--) {
    int u, v, num;
    cin >> u >> v >> num;
    vector<int> nodes = lca(u, v, num);
    int size_node = nodes.size();
    cout << nodes.size() << " ";
    for (int i : nodes) cout << i << " ";
    cout << "\n";
  }
  return 0;
}